Unit 6 · Lesson 2 of 6
Factoring trinomials \(x^2 + bx + c\)
You'll be able to: factor a trinomial that starts with \(x^2\) into two binomials, or say that it is prime.
Step 1 of 5Warm-up
Two quick questions from earlier lessons. Answer from memory first. Pulling it out of your head is what makes it stick.
Factor out the GCF: \(6x^2 - 15x\).
Numbers 6 and 15: GCF 3. Lowest power of \(x\): \(x\). GCF \(= 3x\).
\(\dfrac{6x^2}{3x} = 2x\) and \(\dfrac{-15x}{3x} = -5\).
Answer: \(3x(2x - 5)\).
Compute \((-4)(-9)\).
\(4 \cdot 9 = 36\). Same signs, so the product is positive.
Answer: \(36\).
Step 2 of 5The idea
Multiply \((x + 2)(x + 5)\) and you get \(x^2 + 7x + 10\). Look where the numbers went: \(2 \cdot 5 = 10\) is the last term, and \(2 + 5 = 7\) is the middle coefficient.
Factoring a trinomial runs that backward. You look for the two numbers.
Rule. \(x^2 + bx + c = (x + m)(x + n)\), where \(m \cdot n = c\) and \(m + n = b\).
Sign patterns. If \(c\) is positive, \(m\) and \(n\) have the same sign as \(b\). If \(c\) is negative, they have opposite signs, and the bigger one has the sign of \(b\).
Two variables. \(x^2 + bxy + cy^2 = (x + my)(x + ny)\), with the same two numbers.
No pair works? Then the trinomial is a prime polynomial: it cannot be factored using integers.
List the factor pairs of \(c\) in order (1 and \(c\), 2 and \(c \div 2\), ...) so you don't miss one. Then check your answer by multiplying it back out with FOIL.
Why this shows up on the placement test: factoring trinomials is the most-used factoring skill; you need it to solve quadratics and to simplify rational expressions.
Step 3 of 5Worked examples
Example 1: \(x^2 + 7x + 12\)
- Need: multiply to \(12\), add to \(7\). Both positive. \(c\) and \(b\) are both positive.
- Pairs of 12: \(1, 12\) (sum 13); \(2, 6\) (sum 8); \(3, 4\) (sum 7). Found it.
- Answer: \((x + 3)(x + 4)\). Check: \(x^2 + 4x + 3x + 12 = x^2 + 7x + 12\).
Example 2: \(x^2 - 2x - 15\), then \(x^2 + 3x + 5\)
- Need: multiply to \(-15\), add to \(-2\). Opposite signs, and the bigger one is negative.
- \(3 \cdot (-5) = -15\) and \(3 + (-5) = -2\). Answer: \((x + 3)(x - 5)\).
- For \(x^2 + 3x + 5\): multiply to 5, add to 3. Both positive. The only pair is \(1, 5\), with sum 6.
- No pair works, so \(x^2 + 3x + 5\) is prime. "Prime" is a real answer, not a failure.
Step 4 of 5Practice
Type your answer and press Check. If you miss twice, open the worked solution. Then mark how it went. Type factors with parentheses, like (x + 2)(x - 3).
Factor \(x^2 + 8x + 15\).
Multiply to 15, add to 8, both positive. Pairs: \(1, 15\) (sum 16); \(3, 5\) (sum 8).
Answer: \((x + 3)(x + 5)\).
Factor \(x^2 - 9x + 14\).
Multiply to \(+14\), add to \(-9\): both negative. Pairs: \(-1, -14\) (sum \(-15\)); \(-2, -7\) (sum \(-9\)).
Answer: \((x - 2)(x - 7)\).
Factor \(x^2 + 4x - 21\).
Multiply to \(-21\), add to \(+4\): opposite signs, bigger one positive.
\(7 \cdot (-3) = -21\) and \(7 + (-3) = 4\).
Answer: \((x + 7)(x - 3)\).
Factor \(x^2 - x - 30\).
The middle term \(-x\) means \(b = -1\). Multiply to \(-30\), add to \(-1\): opposite signs, bigger one negative.
\(-6 \cdot 5 = -30\) and \(-6 + 5 = -1\).
Answer: \((x - 6)(x + 5)\).
Factor \(x^2 - 5xy + 6y^2\).
Same method with the numbers 6 and \(-5\): multiply to 6, add to \(-5\). Both negative: \(-2, -3\).
Put a \(y\) with each number: \((x - 2y)(x - 3y)\).
Check: \(x^2 - 3xy - 2xy + 6y^2 = x^2 - 5xy + 6y^2\). Answer: \((x - 2y)(x - 3y)\).
Factor \(x^2 + 2x - 48\).
Multiply to \(-48\), add to \(+2\): opposite signs, bigger one positive. Pairs of 48: \(1, 48\); \(2, 24\); \(3, 16\); \(4, 12\); \(6, 8\).
\(8 \cdot (-6) = -48\) and \(8 + (-6) = 2\).
Answer: \((x + 8)(x - 6)\).
Which trinomial is prime (cannot be factored using integers)?
- \(x^2 + 6x + 8\)
- \(x^2 + 2x - 8\)
- \(x^2 + 5x + 8\)
- \(x^2 - 7x - 8\)
(a) \(2 \cdot 4 = 8\), \(2 + 4 = 6\): \((x + 2)(x + 4)\).
(b) \(4 \cdot (-2) = -8\), \(4 + (-2) = 2\): \((x + 4)(x - 2)\).
(c) Multiply to \(+8\), add to \(+5\). Pairs: \(1, 8\) (sum 9); \(2, 4\) (sum 6). Nothing gives 5, so it is prime. This one.
(d) \(-8 \cdot 1 = -8\), \(-8 + 1 = -7\): \((x - 8)(x + 1)\).
A rectangle has area \(x^2 + 11x + 24\) square feet. Its width is \(x + 3\) feet. Write an expression for its length.
Area = width \(\times\) length, so factor the area and look for \((x + 3)\).
Multiply to 24, add to 11: \(3 \cdot 8 = 24\) and \(3 + 8 = 11\). So \(x^2 + 11x + 24 = (x + 3)(x + 8)\).
Answer: the length is \(x + 8\) feet.
Step 5 of 5Wrap-up
You can now factor \(x^2 + bx + c\) by finding two numbers that multiply to \(c\) and add to \(b\), use the signs to narrow the search, and recognize a prime trinomial.
Read or watch next: OpenStax Intermediate Algebra 2e, 6.2 Factor Trinomials (free textbook, with more practice at the end of the section).
Also good: Khan Academy: Algebra 1: Factoring quadratics intro (short videos). Quick rules: Factoring cheat sheet.
Something unclear? Ask your teacher bot a follow-up question, for example: "How do I know which of my two numbers gets the minus sign?" It knows this course and can explain things another way.